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Question Number 167167 by cortano1 last updated on 08/Mar/22

$$\mathrm{Find}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{function}$$$$\mathrm{f}\left(\mathrm{x}\right)=2\mathrm{x}-\sqrt{\mathrm{x}+1}-\sqrt{{\mathrm{x}}^2-1}$$

Answered by MJS_new last updated on 08/Mar/22

$$f(-1)=-2$$$$f\mathrm{has}\mathrm{a}\mathrm{singularity}\mathrm{at}x=-1$$

Commented by cortano1 last updated on 08/Mar/22

$$\mathrm{not}\mathrm{minimum}\mathrm{value}=\frac{1}{4}?$$

Commented by cortano1 last updated on 08/Mar/22

$$\mathrm{when}\mathrm{x}=\frac{5}{4}$$

Commented by MJS_new last updated on 08/Mar/22

$$\mathrm{yes}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{local}\mathrm{minimum}\mathrm{at}x=\frac{5}{4}$$$$\mathrm{but}\mathrm{the}\mathrm{absolute}\mathrm{minimum}\mathrm{is}\mathrm{at}x=-1$$

Commented by cortano1 last updated on 09/Mar/22

$$\mathrm{no}\mathrm{sir}\mathrm{since}{\mathrm{D}}_{\mathrm{f}}\mathrm{is}\mathrm{x}⩾1$$

Commented by MJS_new last updated on 09/Mar/22

$$\mathrm{no}.{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{defined}\mathrm{for}x=-1\vee x⩾1$$$$f(-1)=2\times (-1)-\sqrt{-1+1}-\sqrt{{(-1)}^2-1}=$$$$=-2-\sqrt{0}-\sqrt{0}=-2$$

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